262 THE DUPLICATION OF THE CUBE A 
N ow MID = BM. MA + DA\ ai 
while, by (1), FK 2 = BK. KG + OF 2 ; dl 
therefore BM. if A + DA 2 = BK . KG + GF'\ tl 
But DA — GF; therefore BM. MA — BK. KG. al 
Therefore GK: MA = BM: BK L 
= LG: GK; 
F 
while, at the same time, fh¥: J5ff = if A : AX. ^ 
Therefore LG:GK = GK: MA = MA : AL, t 
or AB-.GK = GK: i/A = ifA : f?C. 
1 
{6) Apollonius, Heron, Philon of Byzantium. 
I give these solutions* together because they really amount 
to the same thing. 1 
Let AB, AG, placed at right angles, be the two given straight 
lines. Complete the rectangle ABDG, and let E be the point 
at which the diagonals bisect one another. 
Then a circle with centre E and radius EB will circumscribe 
the rectangle ABDG. 
Now (Apollonius) draw with centre E a circle cutting 
AB, AC produced in F, G but such that F, D, G are in one 
straight line. 
Or (Heron) place a ruler so that its edge passes through D, 
1 Heron’s solution is given in his Mechanics (i. 11) and Belopoeica, and is 
reproduced by Pappus (hi, pp. 62-4) as well as by Eutocius (loc. cit.).